The Characteristic Equation

Eigenvalues are the numbers $\lambda$ that make $A - \lambda I$ singular. Determinants detect singularity, so they turn the search for eigenvalues into solving one polynomial equation.

Key idea

$\lambda$ is an eigenvalue of $A$ exactly when $\det(A - \lambda I) = 0$, the characteristic equation. For an $n \times n$ matrix, $\det(A - \lambda I)$ is a polynomial of degree $n$ in $\lambda$, the characteristic polynomial.

Why it works

By the Invertible Matrix Theorem, $(A - \lambda I)\mathbf{v} = \mathbf{0}$ has a nonzero solution exactly when $A - \lambda I$ is not invertible, exactly when its determinant is $0$. Each entry of $A - \lambda I$ is a number or a number minus $\lambda$, and a determinant is a sum of products of $n$ entries, so the result is a polynomial in $\lambda$ of degree $n$.

Worked example: A 2 × 2 matrix

Find the eigenvalues of $A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$.

$A - \lambda I = \begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix}$.
$\det(A - \lambda I)$ $= (4 - \lambda)(3 - \lambda) - (1)(2)$ $= \lambda^{2} - 7\lambda + 10$.
Factor: $(\lambda - 5)(\lambda - 2) = 0$, so $\lambda = 5$ or $\lambda = 2$.
Check: the eigenvalues add to $7 = 4 + 3$ (the trace) and multiply to $10 = \det A$.

Answer: $5$ and $2$

For any $2 \times 2$ matrix the polynomial is $\lambda^{2} - (\operatorname{tr} A)\lambda + \det A$, where the trace $\operatorname{tr} A$ is the sum of the diagonal entries. In general, the eigenvalues (counted with multiplicity) add up to the trace and multiply to the determinant: a free check on your arithmetic.

Worked example: A 3 × 3 matrix

Find the eigenvalues of $B = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & -1 \\ 0 & 2 & 0 \end{bmatrix}$.

Expand $\det(B - \lambda I)$ along row 1, which has two zeros: $$(2 - \lambda)\begin{vmatrix} 3 - \lambda & -1 \\ 2 & -\lambda \end{vmatrix}.$$
The $2 \times 2$ part: $(3 - \lambda)(-\lambda) + 2$ $= \lambda^{2} - 3\lambda + 2$ $= (\lambda - 1)(\lambda - 2)$.
So $\det(B - \lambda I)$ $= (2 - \lambda)(\lambda - 1)(\lambda - 2)$ $= -(\lambda - 2)^{2}(\lambda - 1)$.
Eigenvalues: $2$ (a double root) and $1$. Check: $2 + 2 + 1 = 5 = \operatorname{tr} B$ and $2 \cdot 2 \cdot 1 = 4 = \det B$.

Answer: $2, 2, 1$

The number of times a root repeats is its algebraic multiplicity; here $\lambda = 2$ has multiplicity 2. For a triangular matrix, $A - \lambda I$ is triangular too, so its determinant is the product of the diagonal entries: the eigenvalues of a triangular matrix are its diagonal entries.

Common mistake

Subtracting $\lambda$ from every entry instead of only the diagonal. $\lambda I$ has $\lambda$ on the diagonal and $0$ elsewhere, so the off-diagonal entries of $A - \lambda I$ are the same as those of $A$. Quick check: the eigenvalues should add up to the trace.

This is $A - 5I$ for the first example. Its determinant is $0$: it squashes the plane onto a line, so some nonzero vector (an eigenvector of $A$) is sent to $\mathbf{0}$.

A real polynomial can have complex roots, so a real matrix can have complex eigenvalues (a rotation, for example). We return to these in complex eigenvalues. Also, row reducing $A$ does not preserve eigenvalues; always work with $\det(A - \lambda I)$ for the original matrix.

The Characteristic Equation

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